Question: Simplify the following expression and state the condition under which the simplification is valid: $z = \dfrac{n^2 + 4n}{n^2 + 12n + 32}$
First factor the expressions in the numerator and denominator. $ \dfrac{n^2 + 4n}{n^2 + 12n + 32} = \dfrac{(n)(n + 4)}{(n + 8)(n + 4)} $ Notice that the term $(n + 4)$ appears in both the numerator and denominator. Dividing both the numerator and denominator by $(n + 4)$ gives: $z = \dfrac{n}{n + 8}$ Since we divided by $(n + 4)$, $n \neq -4$. $z = \dfrac{n}{n + 8}; \space n \neq -4$